Please help solve this

an airplane takes off 12.5 miles south of a city and flies due north at a constant speed of 170 miles per hour. What is the plan

saveliy_v [14]

75 miles/hr* (45/60) hours=56.25. Initial position is -12.5. Add -12.5 to 56.25 which results in 43.75 miles due north.

6 0

4 years ago

If a polyhedron has 5 faces and 7 vertices, then it must have how many edges? Please show step-by-step work

Juli2301 [7.4K]

Answer:

10 edges

Step-by-step explanation:

we know that

The Euler's formula state that: In a polyhedron, the number of vertices, minus the number of edges, plus the number of faces, is equal to two

$V - E + F = 2$

in this problem we have

$V=7\\E=?\\F=5$

substitute the given values

$7 - E + 5 = 2$

solve for E

Combine like terms in the left side

$12 - E = 2$

subtract 12 both sides

$- E = -10$

Multiply by -1 both sides

$E =10$

8 0

3 years ago

What is the value of x? Enter your answer in the box.

Lapatulllka [165]

We can see that

there are six sides

so, n=6

so, firstly we will find total angle

total angle =(n-2)*180

total angle =(6-2)*180

total angle =720

so, sum of all angles must be 720

so, we get

$x+120+100+128+133+112=720$

now, we can solve for x

$x+593=720$

$x=127$ .............Answer

5 0

3 years ago

Find 5 consecutive whole numbers if it is known that the sum of the squares of the first 3 numbers is equal to the sum of the sq

lora16 [44]

Let the first number of the five we are looking for equal n.

Because the numbers are consecutive, the next four can be expressed as (n+1), (n+2), (n+3) and (n+4).

Our series of 5 numbers is therefore n,(n+1),(n+2),(n+3),(n+4).

We are told the sum of the squares of the first 3 is equal to the sum of the squares of the last 2 therefore:
n² + (n+1)² + (n+2)² = (n+3)² + (n+4)²

expand all the brackets to give
n² + n² + 2n + 1 + n² + 4n + 4 = n² + 6n + 9 + n² + 8n + 16
3n²+6n+5 = 2n²+14n+25
n²-8n-20=0

factorising this gives us
(n+2)(n-10) = 0 so the solutions are n=10 and n=-2

That means that n,(n+1),(n+2),(n+3),(n+4) could be [-2, -1, 0, 1 and 2] or [10, 11, 12, 13 and 14].

Since the question asks for whole numbers, and negative numbers are not classed as whole numbers, the numbers we want must be 10, 11, 12, 13 and 14.

You can check this on a calculator by doing 10² + 11² + 12² (=365) and 13² + 14² (=365).

3 0

3 years ago

7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that

kompoz [17]

Answer:

a)

The probability that a single score drawn at random will be greater than 110

P( X > 110) = 0.1587

b)

The probability that a sample of 25 scores will have a mean greater than 105

P( x> 105) = 0.0062

c)

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

Step-by-step explanation:

a)

Given mean of the Normal distribution 'μ' = 100

Given standard deviation of the Normal distribution 'σ' = 10

a)

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

$Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1$

The probability that a single score drawn at random will be greater than 110

P( X > 110) = P( Z >1)

= 1 - P( Z < 1)

= 1 - ( 0.5 +A(1))

= 0.5 - A(1)

= 0.5 -0.3413

= 0.1587

b)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5$

The probability that a single score drawn at random will be greater than 110

P( x> 105) = P( z > 2.5)

= 1 - P( Z< 2.5)

= 1 - ( 0.5 + A( 2.5))

= 0.5 - A ( 2.5)

= 0.5 - 0.4938

= 0.0062

The probability that a single score drawn at random will be greater than 105

P( x> 105) = 0.0062

c)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4$

The probability that a single score drawn at random will have a mean greater than 105

P( x> 105) = P( z > 4)

= 1 - P( Z< 4)

= 1 - ( 0.5 + A( 4))

= 0.5 - A ( 4)

= 0.5 - 0.498

= 0.002

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

Let x₁ = 95

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2$

Let x₂ = 105

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2$

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

= P(z≤2) - P(z≤-2)

= 0.5 + A( 2) - ( 0.5 - A( -2))

= A( 2) + A(-2) (∵A(-2) =A(2)

= A( 2) + A(2)

= 2 × A(2)

= 2×0.4772

= 0.9544

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

7 0

3 years ago